# Comparison of Exact Constants in Inequalities for Derivatives of Functions Defined on the Real Axis and a Circle

• V. F. Babenko
• V. A. Kofanov
• S. A. Pichugov Днепропетр. нац. ун-т ж.-д. трансп.

### Abstract

We investigate the relationship between the constants K(R) and K(T), where $K\left( G \right) = K_{k,r} \left( {G;q,p,s;\alpha } \right): = \mathop {\mathop {\sup }\limits_{x \in L_{p,s}^r \left( G \right)} }\limits_{x^{(r)} \ne 0} \frac{{\left\| {x^{\left( k \right)} } \right\|_{L_q \left( G \right)} }}{{\left\| x \right\|_{L_q \left( G \right)}^\alpha \left\| {x^{\left( r \right)} } \right\|_{L_s \left( G \right)}^{1 - \alpha } }}$ is the exact constant in the Kolmogorov inequality, R is the real axis, T is a unit circle, $$L_{p,s}^r (G)$$ is the set of functions xL p(G) such that x (r)L s(G), q, p, s ∈ [1, ∞], k, rN, k < r, We prove that if $$\frac{r - k + 1/q - 1/s}{r + 1/q - 1/s} = 1 - k/r$$ thenK(R) = K(T),but if $$\frac{r - k + 1/q - 1/s}{r + 1/q - 1/s} < 1 - k/r$$ thenK(R) ≤ K(T); moreover, the last inequality can be an equality as well as a strict inequality. As a corollary, we obtain new exact Kolmogorov-type inequalities on the real axis.
Published
25.05.2003
How to Cite
Babenko, V. F., V. A. Kofanov, and S. A. Pichugov. “Comparison of Exact Constants in Inequalities for Derivatives of Functions Defined on the Real Axis and a Circle”. Ukrains’kyi Matematychnyi Zhurnal, Vol. 55, no. 5, May 2003, pp. 579-8, https://umj.imath.kiev.ua/index.php/umj/article/view/3934.
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Section
Research articles